Determine the number of solutions to
\[2\sin^3 x - 5 \sin^2 x + 2 \sin x = 0\]in the range $0 \le x \le 2 \pi.$
Solution: The given equation factors as
\[\sin x (2 \sin x - 1)(\sin x - 2) = 0,\]so $\sin x = 0,$ $\sin x = \frac{1}{2},$ or $\sin x = 2.$

The solutions to $\sin x = 0$ are $x = 0,$ $x = \pi,$ and $x = 2 \pi.$

The solutions to $\sin x = \frac{1}{2}$ are $x = \frac{\pi}{6}$ and $x = \frac{5 \pi}{6}.$

The equation $\sin x = 2$ has no solutions.

Thus, the solutions are $0,$ $\pi,$ $2 \pi,$ $\frac{\pi}{6},$ and $\frac{5 \pi}{6},$ for a total of $\boxed{5}$ solutions.